In this video I want to derive
the current-voltage relationship for a p-n diode.
And I want to derive it for two cases.
One is when there's no light shining
on the diode, and that's called a dark IV relationship.
And then I want to contrast it to the case where there's light shining
on the diode and that's in fact
called very appropriately called light JV characteristics.
So let's start first
with the case where we are in the dark for the lower light, shining on this p-n diode.
And then I can apply a voltage across this p-n diode.
So I can apply a voltage across this p-n diode.
And that voltage essentially results in a reduction
of the built-in voltage across this diode.
So let's say the built-in voltage I'm applying a voltage of VA.
And the built-in
voltage was Vbi.
So, now the potential difference between this p and n region,
to Vbi minus VA.
And all of that applied voltage
it essentially appears across our depletion region.
So assuming that these p and n region
have a decent conductivity so all that potential
difference will occur across this depletion region, where
there is a higher resistance or no carrier presence,
all this applied voltage occurs across
this depletion region. And that results in splitting of these
Fermi levels into two of these quasi-Fermi leves
. Denoted by EFP Which are used to represent
the carrier concentration of holes, and EFn, which is used to represent the
carrier concentration of electrons. So now, I mean in thinking about it
What I am doing is essentially there
was this barrier for present for the preventing
these diffusions of electrons from the
n-type semiconductor entering into the p-type semiconductor.
But now since I've reduced this barrier, these electrons can
essentially now go and go into the p-type semiconductor.
Similarly these holes which are present in the p-type
semiconductor, they now see a reduced barrier to come into the n-type semiconductor.
So there is a flow of current which
results as as a consequence of that,
and that can be represent essentially like a like a rectifier relationship
where this current density can be related to the applied voltage.
You link this exponential formula, where a, it a, varies exponentially with
the applied voltage and it's also highly rectifying in nature.
Where just J0 it essentially depends upon a lot of things.
So it's given by this formula which I'm stating over here.
And it can be derived by applying the drift-
diffusion equation and the continuity equation across this diode.
So this formula essentially says that my my
diode current at you know, this J01,
its relation to the, it's related to the diffusion constant in
the p and n regions and it's related to the diffusion length,
it's deeply related to the doping density
and very importantly, it's related to the square
of the intrinsic carrier concentration.
So it's related to the square of the intrinsic carrier concentration.
And if I, if I chart this IV relationship,
it would essentially look like a ideal rectifier relationship.
So when I apply, put voltage in a
positive direction, so if I forward bias this diode,
there will be an exponential increase there'll be exponential increase in
the current.
And when I go into the reverse biased region, what I'll see is that my current
it it turns negative, and it does not essentially vary much with the voltage.
So in fact as this VA here becomes
very negative, this VA here becomes very negative,
then my J will essentially, you know, it will
become equal to minus J01, and that's called reverse
saturation current.
So this is how my this is how my IV relationship would look like.
If I apply a positive voltage, there's a large, there's a large change in my
current, versus if I apply a negative
voltage, there's a small change in my current.
So now let's move to the case where we in fact apply light on this p-n junction.
So this,
this, when I do apply light on this p-n junction.
And let's consider the case where the light, the sunlight
is shining from this direction which is incoming from the p side.
So, this light will get, these photons will get absorbed inside these.
Inside these p and n regions.
And they'll they'll simply result in the production,
result in this creation of these electrons and hole pairs.
And this can be represented by this generation rate which would
decrease as the light gets as the light traverses through this
p-n junction, as the light gets absorbed so the generation rate would decrease.
So now the electron and whole pairs which are generated inside
this depletion region,
and this is I'm still considering the case where I'm applying this voltage VA.
So you saw that these electrons and holes earlier were flowing in this direction.
But now when I'm shining light on this diode, these electrons and holes pair
which are generated in this In this depletion region,
they would essentially be swiped out, so this electron
over here would get swiped out into the n region.
Because these electrons like to travel downhill.
Similarly, these holes that are generated inside the depletion region, they would get
swiped out into the p region and where they would be collected as a current.
So, it would result in this result in this generation current, which
is in fact opposite in polarity as compared to the other case.
So I can say that my generation current, due to this, due to this light
shining on my diode, I can represent it by
essentially integrating the generation rate over my depletion region.
So I can integrate this generation rate, which
is a function of distance, over this depletion region.
And that would be essentially the current which is generated
in opposite direction, due to the presence of this light.
So let me write this down, so this generation
current can now be written down by this formula,
where it's integrated from the beginning of this depletion region.
So let's say this is my zero coordinate.
So it's integrated from minus xp to xn. And now my, I can write my
overall current as a superposition of the two currents flowing before
and the current which was flowing before the,
any light was shining on this diode. And then superimpose this
with this current which is now flowing due to the light shining.
So my new IV relationship or my new JV relationship for my diode can essentially
be represented by this formula. Where it's a superposition of what
was happening before, in the dark.
And now it's superimposed with this other
current, which is flowing in the opposite direction.
So now if I consider the case when I'm applying no
voltage or I'm applying zero voltage across my my p-n junction,
I'm essentially shorting my p-n junction,
then the current that I would expect is I can call this as the short
circuit current.
And if I put V applied equal to 0 in this formula, so this term would vanish and the
current that I would get is essentially equal to is equal to this this value.
So I can I can represent this equation in terms of the short circuit current.
as J is equal to J01
exponential function of my applied voltage
minus my, minus my short circuit current, when I'm shining light.
So now my if I go back to this IV relationship that I derived earlier,
so I had mentioned that the, I can represent my diode
IV relationship, and it looks very much like a rectifier.
So if I write the JV relationship of a diode. Let me write it first for
the case of where there is no light applied, and it looks something like this.
So this is a, this is the case when there's a dark IV.
Now if I apply, apply a light, what I'm seeing
essentially is that most of my JV relationship is the same.
Only that I have shifted it downward by this short circuit current.
So now if I consider the case where I have light applied on my junction,
so my IV relationship would essentially look very much the same.
But it would be shifted negatively in, in terms
of the y-axis, by this amount equal to the Jsc.
So it would essentially look something like this.
And so if
I look at this, IV relationship.
The region in which I can extract power out of this
device would be when I operating this, operating this device in this fourth quadrant.
So when I'm operating my device in this fourth quadrant,
so I'm apply a positive voltage while I'm getting a reverse current.
So this would be the case where I can use this, p-n junction as a solar cell.